Browse Questions

# Verify mean value theorem for each of the function $f(x)=\large\frac{1}{4x-1}$ in $[1,4].$

Toolbox:
• Let $f(x)$ be a function defined on $[a,b]$ such that
• (i) it is continuous on $[a,b]$
• (ii) it is differentiable on $(a,b)$
• Then,there exists $c\in (a,b)$ such that $f'(c)=\large\frac{f(b)-f(a)}{b-a}$
Step 1:
Given : $f(x)=\large\frac{1}{4x-1}$ in $[1,4]$
Since for each $x\in [1,4]$,the function attains a unique definite value.
So $f(x)$ is continuous on $[1,4]$
Also $f'(x)=\large\frac{-4}{(4x-1)^2}$ exists for all $x\in (1,4)$.So it is differentiable on $(1,4)$.Hence both the conditions of Lagrange's Mean value theorem are satisfied.
Hence there must be one value of $c\in (1,4)$ such that $f'(c)=\large\frac{f(4)-f(1)}{4-1}$
Step 2:
$f'(x)=\large\frac{-4}{(4x-1)^2}$
Therefore $f'(c)=\large\frac{-4}{(4c-1)^2}$
$f(1)=\large\frac{1}{4\times 1-1}=\frac{1}{3}$
$f(4)=\large\frac{1}{4\times 4-1}=\frac{1}{15}$
Therefore $f'(x)=\large\frac{1/15-1/3}{4-1}=\frac{-4}{45}$
$\Rightarrow \large\frac{-4}{4(-1)^2}=\frac{-4}{45}$
$\Rightarrow 45=4(-1)^2$
$\Rightarrow 16c^2-8c+1=45$
$16c^2-8c-44=0$
On simplifying we get,
$4c^2-2c-22=0$
$2c^2-c-11=0$
Step 3:
Solving for $c$ by applying the quadratic formula,
$C=\large\frac{1\pm\sqrt{1+88}}{4}$
$C=\large\frac{1\pm\sqrt{89}}{4}$
But $\sqrt{89}=9.43$
Neglecting the negative value
$\large\frac{1+9.43}{4}=\frac{10.43}{4}$
$c=2.607$
Clearly the value of $c$ lies between $[1,4]$
Hence Lagrange's Mean value theorem is verified.